Define the following function on ON (The class of ordinals): $f(0)=0$ and for every $\alpha >0$, if $\alpha$ is a successor then $f(\alpha) = \alpha -1$ and if $\alpha$ is limit then, $f(\alpha) = \alpha$.
I get, by this, a regressive function on the class of all ordinals, s.t. for each $\gamma \in ON$, the set $\{ \alpha | f(\alpha) < \gamma \}$ is bounded.
I think this can not be true so:
What is wrong with the function that I have defined?
Thanks
Edit: The reason I ask this is that if this is true then wouldn't it make the following exercise trivial?
$\bf 8.4$ If $X\subseteq\kappa$ is nonstationary, then there exists a regressive function $f$ on $X$ such that $\{\alpha:f(\alpha)\leq\gamma\}$ is bounded for every $\gamma<\kappa$.
[Let $C\cap X=\varnothing$ and let $f(\alpha)=\sup(C\cap\alpha)$.]
The function is not regressive at all on non-successor ordinals. Regressive means $f(\alpha)<\alpha$, but if $\alpha$ is not a successor, then $f(\alpha)=\alpha$.
This shows that the class of successor ordinals is not stationary. Which is what it should be, since the limit ordinals is a closed and unbounded class of ordinals.